Friday, January 30, 2009

Wordplay (e.g., using the language to provoke thought) is a recreational sport that builds a great deal of mind power. Some types of expressions which truly employ wordplay are oxymoron (which we played with yesterday), paradoxes (where two thoughts expressed within a sentence are self-negating), bizarre metaphors (exaggerated comparisons -- which tend to be humorous or particularly memorable) and alliterations (where the first consonant sounds of of several words in a row are the same for enhanced emphasis). Some examples of these last three types of wordplay follow.

Your mission (should you choose to accept it), is to think of at least three additional examples ofyour own to add to my little collection in each category. This is a wonderful distraction when you are waiting in the dentist's office, being audited by the IRS, or watching your sister-in-law's slide show of her family's trip to a water park.

PARADOXES

1. I am a pathological liar -- if you can believe that.

2. I must have told you five million times not to exaggerate.

3. You are every woman in the world to me. (with apologies to the 70's group, Air Supply)

4. We worked 25 hours per day to get the budget prepared.5. The more things change, the more they remain the same.

BIZARRE METAPHORS

1. He ran like a bat out of hell.2. The State's Attorney had all the morality of a frequently-relocated priest.3. She was every bit as pure as the yellow snow.4. Dad had a smile which could curdle milk.5. Cousin Phil was an undertaker who made a living while everyone else died.

ALLITERATIONS

1. Pusillanimous peons for peace.2. Weary and wasted wanderers.3. Fabrications about his fast and furious fencing with foes.4. Whispering words of warmth.5. Methods to mitigate madness.

Have fun with these. By the way, if you come up with some truly great ones, please post them as comments to this blog. I'll give you credit and recognition if we use them in the future.

Thursday, January 29, 2009

Yesterday, you gave your mind a great workout. By the way, the answer to questions #3 is "yes, the number of possibilities and the ease of coming up with words would increase if the numerical sum were increased from 15 to 20." The rest of the questions were not really questionas at all...they were instructions to engage in exercise, in the true Braintenance tradition.

Here are several more to try:

1) How many oxymorons can you think of? An oxymoron is a combination of words which are conflictory, or mutually-negating. Some are very obvious, and some are quite subtle. Some are stereotypical and subjective, and others are cultural. Some examples (a mixture) include:

2) Can you compose a Haiku? A Haiku is a Japanese poetry form which consists of three lines; the first line has five syllables; the second line has seven syllables; and the third line has five syllables. An example would be:

I stand at the door

Watching as you walk way

Your last words haunt me

3) How many ways can you think of to complete the following sentence, given only one minute?

Wednesday, January 28, 2009

Your intelligence grows through idea association -- through simple exercises or challenges where you are required to create something new by putting together information stored in a variety of different "compartments" in your memory banks. The more of this you do, the more you widen and strengthen those intercompartmental pathways so that knowledge can freely flow back and forth between them at greater speeds. Some exercises follow which seem rather simplistic, but which will actually make you use you mind and memory in ways that you may not ordinarily do. Try them! You'll have lots of fun (but tell no one).

1. Find as many words as you can to rhyme with each of the following words, giving yourself no longer than one minute for each one. You may use multiple words to create one rhyme (ex. "Trinidad" and "Half as bad" would be acceptable). Most individuals would struggle to come up with two viable rhymes for a word of more than three syllables...if you can come up with five, you are better than average...if you can come up with ten, you are truly gifted.

Friday, January 23, 2009

Now that you have had the opportunity to enjoy a well-earned day's rest, let's switch the format of our meeting. I offer some solutions before I pose new problems. Variety is an intellectual stimulant (as it is a leading cause for divorce in the US...but then, that's in a different context). Please refer to the last two posts on this blog, as it's generally a good policy to know the questions tha the answers relate to!-------

Question: What would the amount and the dilution be of hyroxyidiotic acid be in bottle #3 at the end of this process? This is a question that can be answered intuitively, without much math at all (phew). In Question #3, we were left with 1 gallon of 20% hydroxyidiotic acid in bottle #3. In Question #4, we added 1 gallon of 10% of hydroxyidiotic acid to bottle #3. In sum, bottle #3 now contains 1 gallon of 20% acid, and 1 gallon of 10% acid, or 2 gallons of 15% hydroxyidiotic acid. If we were to have done the math, we might have used this calculation: (.50) (.20) +(.50) (.10) = .15 , or 15%._________________________________________How many 5-letter combinations can you make of the name OBAMA, where none of those combinations started with an "O"? The computation is simple (but with a twist): for the first letter slot you have 4 possible letters; for the second, 4; for the third, 3: for the fourth, 2; and for the fifth, 1 --- by multiplying 4x4x3x2x1, we get 96 arrangements of letters that do not begin with the letter "O". But what about the arrangements that are redundant because they have two "A"s in them....Hmmm....perhaps you need a bit more time. Since every arrangement of these letters will contain two letter "A"s, and it doesn't matter which A is being used, half of our arrangements will be redundant. What we must then do is eliminate half of the arrangements in order to eliminate these redundancies. If we divide 96 in half, the result will be 48 arrangements, which is our answer.______________________________________Now for a second surprise...there is no problem for today. There is merely a wonderful exercise. Take at least 30 minutes at some time time today, and sample some of the audio/video clips below for mind expansion purposes. Scroll down to EXERCISE YOUR MIND 6: MEDITATION, STIMULATION AND RELAXATION. Enjoy the experience!

Wednesday, January 21, 2009

It is a scientifically documented fact that the exercise of freely associating words (i.e., connecting words that are in some way related -- either as synonyms, opposites, parts of each other, complements of each other, sounds in common, used in the same industry, commonly used in association with other words as part of a phrase) helps increase cognitive function, opens up new neural pathways, builds up greater n-dimensional reasoning, and strengthens memory. Try to find unusual associations, as well as the traditional ones:

1. disgruntled

2. contemplating

3. stainless

4. winding

5. summer

6. animal

7. testing

8. innoculation

9. locomotive

10. trailer

Enjoy these. and remember that there are no right or wrong answers. come up with as many associations for each word as possible -- and bear in mind that each association will have its own underlying rationale. These associations give you a special opportunity to examine the workings of how you think!

Faithfully,

Douglas Castle

Solutions to earlier quizzes follow:

Given: You have the bottles. One of them is a 2-gallon bottle filled with a 40% solution of hxdroxyidiotic acid (I have invented this substance); one is a 2-gallon bottle filled with water; one is a 4-gallon bottle, which is empty. You have access to additional water at a nearby pump.

Find:

1. How would you create 4 gallons of a 20% solution of hydroxyidiotic acid? Simply pour the contents of the first container and the second container into the third container.

2. How would you create 2 gallons of a 20% solution of hydroxyidiotic acid? Take the mixture created in #1, above, and fill one of the empty 2 gallon bottles with it. Fill the second bottle as well...why waste? what if there were an hydroxyidiotic acid shortage?

3. How would you create 1 gallon of a 20% solution of hydroxyidiotic acid? Due to your limited materials, the easiest (albeit imprecise) way of accomplishing this would be to pour half of the contents of either bottle #1 or #2 into bottle #3. The end result would be 1 gallon of 20% hydroxyidiotic acid in the container used (assume that you used container #1), 2 gallons of 20% hydroxyidiotic acid in bottle#2, and 1 gallon of 20% hydroxyidiotic acid in bottle #3.

4. How would you create 2 gallons of a 10% solution of hydroxyidiotic acid? Fill the balance of bottle #1 with water to cut double the dilution (or halve the strength) to 10%.

5. How would you create 1 gallon of a 10% solution of hydroxyidiotic acid? Due to your limited materials, you could simply pour half of the contents of bottle #1 into bottle #3. Bottle #1 would then contain 1 gallon (approximately) of 10%hydroxyidiotic acid.

Question: What would the amount and the dilution be of hyroxyidiotic acid be in bottle #3 at the end of this process?

_________________________________________

How many 5-letter combinations can you make of the name OBAMA, where none of those combinations started with an "O"? The computation is simple (but with a twist): for the first letter slot you have 4 possible letters; for the second, 4; for the third, 3: for the fourth, 2; and for the fifth, 1 --- by multiplying 4x4x3x2x1, we get 96 arrangements of letters that do not begin with the letter "O". But what about the arrangements that are redundant because they have two "A"s in them....Hmmm....perhaps you need a bit more time.

Tuesday, January 20, 2009

Today was the inauguration of the 44th president of the United States Of America. In honor of this much-heralded event, I will post the solution to the last quiz tomorrow.

In the meantime, please scroll on up to find a new Warm-Up exercise: We are happy to present you with the GRE WORD OF THE DAY. We now have a 3rd warm-up exercise before we get into the tough stuff. Please give it a try. And remember to make this site a favorite.

A few moments of Braintenance per day can absolutely create wonders. Do this for yourself!

Faithfully,

Douglas Castle

p.s. How many combinations of 5 letters each can you make from "OBAMA" which do not begin with the letter "O"?

Monday, January 19, 2009

If you combine math with imagination, you get problem-solving ability. The exercise du jour (goodness...have I spoken French?....no, I've written it), requires a bit of imagination and a bit of math. The type of thinking involved is very practical and worth cultivating.

Given: You have the bottles. One of them is a 2-gallon bottle filled with a 40% solution of hxdroxyidiotic acid (I have invented this substance); one is a 2-gallon bottle filled with water; one is a 4-gallon bottle, which is empty. You have access to additional water at a nearby pump.

Find:1. How would you create 4 gallons of a 20% solution of hydroxyidiotic acid?2. How would you create 2 gallons of a 20% solution of hydroxyidiotic acid?3. How would you create 1 gallon of a 20% solution of hydroxyidiotic acid?4. How would you create 2 gallons of a 10% solution of hydroxyidiotic acid?5. How would you create 1 gallon of a 10% solution of hydroxyidiotic acid?

Good luck!

Faithfully,

Douglas Castle______________________________________________________Solution (no pun intended in light of the foregoing) to the last exercise:

If a person takes one step back for every two steps forward, it means that he or she only advances one step ahead for each two steps taken. This is actually a formula. If he or she were to take 20 steps in total, he or she would only have gained ten steps (i.e., be ten steps ahead) after having taken a total of 2o steps.

Friday, January 16, 2009

If an individual tends to take two steps forward, and then one step back, how many steps will he have advanced after having made a total of 20 steps (including both forward and backward)?

Faithfully,

Douglas Castle

______________________________________________________Commentary on the last quiz:

You should learn at least one new word every day.You should look up words which you do not know, instead of defining them "in context".When you learn a new word, try to find several synonyms for it.When you learn a new word, try to find several antonyms for it.When you learn a new word, be certain to use it in at least five sentences that day._______________________________________________________

Thursday, January 15, 2009

You might wish to go to www.dictionary.com and use the Thesarus Feature to come up with at least 5 synonyms and 5 antonyms for the words which follow. In fact, you might wish to look some of these words up in the dictionary before you begin your work:

Wednesday, January 14, 2009

Douglas Castle_______________________________________________Here are some answers to the last quiz:

1. What is the probability that no lightbulbs will go out?

The probability that any one lightbulb will not go out is .95 (which is 1.00 - .05). The probability that ten out of ten lightbulbs will survive the night would be (.95) (.95) (.95) (.95) (.95) (.95) (.95) (.95) (.95) (.95) = .60 or 60%

2. What is the probability that any one lightbulb will go out?

The probability that any one of the lightbulbs (it doesn’t matter which) will go out in the group of ten would be found by dividing .05 [the probability that any one single lightbulb goes out] by .60 [the probability that none of the lightbulbs go out], which would equal .08 or 8%. The mistake that many people make in working with this problem is that they will multiply (.95) times itself nine times, and then multiply that result by .05. This does not work, because each lightbulb operates independently of the others. The answer that you would obtain that way would be smaller than .05, which clearly would not make sense – intuitively, you realize that the answer doesn’t make sense if it is less than .05. Intuition is an important tool!

3. What is the probability that any two lightbulbs will go out (together)?

The probability that any two lightbulbs would go out together, i.e., jointly, would be (.05) (.05) or .0025, or .25%. Using the same reasoning as in the solution to the second problem above, we would obtain our answer by dividing .0025 by .60. The solution would be .000416 or .0416%. Intuitively, this also makes sense because the probability of two lightbulbs going out together should be smaller than the probability of only one lightbulb going out in the whole group.

4. What is the probability that up to three lightbulbs will go out?

This question is asking for the sum of the probabilities that: one lightbulb will go out alone; that two lightbulbs will go out together; and that three lightbulbs will go out together. To solve this one, we add the solution to the second problem (.08), to the solution to the third problem (.025), and then add (.05) (.05) (.05)/ (.60) [the probability of three bulbs going out together], and we get: .08 + .000416 + .000208 = .080624 or 8.06%. Intuitively this makes sense because the additive probabilities of all three cases would have to be a number greater than any one of the individual numbers.

5. What is the probability that either three or four lightbulbs will go out?

I am feeling exhausted and even a bit cantankerous. I’ll let you figure this one out yourself, while I have coffee. :)

Tuesday, January 13, 2009

Today's exercise is fairly easy, but you must read each question carefully in order to understand precisely what is being asked of you! Again: Read Carefully.

A room is lit by ten independently wired (and powered) light bulbs. The probability of a bulb going out during any given night is .05 or 5%. The questions which follow involve one particular night.

1. What is the probability that no lightbulbs will go out?

2. What is the probability that any one lightbulb will go out?

3. What is the probability that any two lightbulbs will go out (together)?

4. What is the probability that up to three lightbulbs will go out?

5. What is the probability that either three or four lightbulbs will go out?

Enjoy this exercise.

Before we get to our solutions to the last quiz, here's an article of interest (reprinted from INTERNAL ENERGY PLUS) that might provide you will an enjoyable stress-buster, as well as an introduction to the amazing power of the mind to manifest changes in physiology and mood.

Dear Friends:

As many of you know, I invest a great deal of time discussing some of the less-known but powerful Modalities (i.e., special tools and techniques) of INTERNAL ENERGY PLUS. One of them is Entrainment (re-programming your emotions through direct signalling to the brain), and another is Deliberate Breathing. These are the two which you can use, right now, to create an amazing, soothing, re-energizing effect.

Here's a stress-reduction exercise that works miraculously, and in less than five minutes:

1. Sit down comfortably in a quiet place. Close your eyes. Breathe slowly, deeply and deliberately for twenty breaths. If you do this properly, you will visualize negativity leaving your mouth like a cloud of ashes, and golden light entering your body with every refreshing, rich inhalation. Your muscles will relax. You will feel a bit lightheaded. Just focus on your breathing. Open your eyes, and stretch your back (still seated).

2. Click on the following audio clip (at a fairly high volume), close your eyes, and visualize a golden light running its course, up and down, repeatedly, from the base of your spine to the very top and center of your head...almost like a warming internal massage, clearing an energy pathway through your body.

_________________________________________________________And now, the answers to the last quiz! The first two of these questions could be answered with good estimates and at warp speed by simply using the "Rule of 72". The rule states that "if you divide the annual interest rate (expressed as a percentage, and NOT A DECIMAL EXPRESSION) into 72, you will find out how many years it will take for your money to double". The rule further states that "if you divide the number of years into 72, you will find out the interest rate required in order to double your money." It's that simple! 1. How long will it take to double your money at 14.4% interest compounded per year? If we divide 72 by 14.4, our answer will be approximately 5 years.

2. At what interest rate, compounded per year, will it take to double an initial investment of $1,000.00 in 5 years? If we divide 72 by 5 years, our answer will be approximately 14.4%.

3. What sum must we initially deposit in order to accumulate the sum of $3,000 in 10 years at an annual compounded rate of 8.00%?Gosh, I feel moody today. Even a bit surly. I won't give you the answer just yet. Instead, I'll provide a hint: At the end of 1 year, the starting amount ("X") will have grown to X(1.08). At the end of two years, X will have grown to X(1.08)(1.08). Do you see a pattern???Stay tuned!

Saturday, January 10, 2009

1. How long will it take to double your money at 14.4% interest compounded per year?

2. At what interest rate, compounded per year, will it take to double an initial investment of $1,000.00 per year in 5 years?

3. What sum must we initially deposit in order to accumulate the sum of $3,000 in 10 years at an annual compounded rate of 8.00%?

Good Luck!

Faithfully,

Douglas Castle_________________________________________________________________Answers to the last quiz follow:

Bob is 7 years old and Ray is 11 years old. The technique to solve this problem reuires some simple algebraic substitution and a knowledge of simultaneous equations. Hint: Ray's age is actually Bob's age + 4. This is really your easiest starting point.

Wednesday, January 7, 2009

1. Bob and Ray are brothers. Bob is younger.2. They are 4 years apart in terms of their respective ages.3. Their combined ages, in three years, will be 24.4. The product (by multiplication) of their ages in three years will be 140.

Q: What is Ray's age now? What is Bob's age now?

Faithfully,

Douglas Castle_________________________________________________________________If you would like the answers to the last quiz, you will have to avail yourself of a dictionary. By the way, the word most often used improperly is "notoriety." It does not mean "fame"...it means "to be known for ill deeds, to be maligned, the state of being notorious". When your buddy says, "I want to make a hit record and achieve notoriety", he or she is misusing the term, unless the hit record is going to be very controversial or of poor quality.

Friday, January 2, 2009

There's no quiz for today. Your exercise is to invest some time in planning how to make 2009 the best year in your life's experience. I am hopeful that you will be dedicate some of this time to getting to know your inner-selves better, and to resolving to become brighter, better thinkers, and greater contributors to the Collective Body of Human Knowledge and Experience.

Faithfully,

Douglas Castle____________________________________________________Answers to the last two quizzes are posted below:

Assume (although we should not assume too often) that we have two men and two women whom we wish to seat in a row:

1) How many ways can these four people be arranged? In 24 different ways. The reasoning: There are 4 people to choose from to occupy the first seat; 3 people left to occupy the second; 2 to occupy the third; and 1 to occupy the last. 4x3x2x1 = 24 different possible arrangements.

2) How many ways can these people be arranged if they must be seated with people of the opposite sex seated next to eachother? There are four configurations for seating, which are MFMF, FMFM, FMMF and MFFM (where M = male and F = female). The computations for each configuration are 2x2x1x1 (=4 arrangments), 2x2x1x1 (= 4 arrangements), 2x2x1x1 (= 4 arrangements), and 2x2x1x1 (=4 arrangements). If we add these together, we get 16 different ways.

3) How many ways can these people be arranged if they must be positioned with people of the same sex next to each other? There are four configurations for seating, which are MMFF, FFMM, FMMF and MFFM. Using the same problem-solving strategy as above, we get a total of 16 different ways.

Many people instantly solved 1) and 2), and thought that the answer to 3) could be obtained by simply subtracting 2) from 1). This is wrong because both 2) and 3) share two configurations which satisfy the rules of each arrangement. Another tricky (counter-intuitive) issue is the notion that 16 ways plus 16 ways = 32 ways. Obviously 32 is greater than 24. But when we think about it, we realize that this is indeed sensible, because of the fact that 2) includes some of the same arrangements as in 3), so there will have to be an element of double-counting._____________________________________________________________Our remaining problem from the previous day's quiz:

3. What is the probability that, if the first ball's color was not identified and eliminated from the game, that the next ball that you selected would blue?

Yep. A trick question. The answer would be the same as in question 1, above, because you are merely performing the same operation over again with the same number of balls. However...if you read the question thinking that a ball with an unidentified color was eliminated and not replaced, the answer is more complicated. It is what my math tutor used to call a "conditional probability problem". It poses a challenge.

If the eliminated ball were red, the number of blue remaining would be 30, with 19 red remaining; but if the eliminated ball were blue, the number of blue remaining would be 29, with 20 red left. In either case, the number of balls left over would be 49. What to do?

If a blue had been eliminated, the new probability of a blue being chosen would be 29/49 (.59); if a red had been eliminated, the new probability of blue being chosen would be 30/49 (.61). Either one of two things has occurred, and we do not know which...but we do know the likelihood (probability) of each one of these events. We know that the first one (a blue ball being chosen) was 30/50 (or .60), and that the second one (a red ball being chosen) was 20/50 (or .40).The computation would be as follows (.60)x(.59) + (.40)x(.61) = .354 [the probability of a blue being chosen after a blue was eliminated from the group] + .244 [the probability of a blue being chosen after a red was eliminated from the group] = .598, or 59.8% . What we have done is to find the probabilities of each possibility (.60 and .40), and multiplied each by its resultant probability of blue being chosen. We added them together because we had to count both of possible ways in which we eliminated a ball from the group, i.e., by picking a blue one first, or by picking a red one first. (phew!)